2007-08-28 11:19:49 · 6 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
like which numbers are divisble to each of themm??
2007-08-28 11:24:07 · update #1
Find the greatest common factor...
Take a number, say 248.
What goes into 248? 4 and 62.
Now, 2 and 2 make four.
(remember 2^2)
So what makes 62? 2 and 31.
Since 31 is prime, that's the end of that.
The factorization of 248 is 2^3 x 31
Do that for each of them...
Then, line them all up.
See which numbers are common (all 4 of them must have it, it doesn't matter what the exponent is)
Then, for each of those common numbers, take the one with the smallest exponent (say there's a 2, a 2^4 and a 2^6. Take the 2 because it has the smallest exponent of 1)
Multiply them all together. TA DA!!!
Sorry if this seems long, once you get the hang of it, it's really easy.
2007-08-28 11:22:09 · answer #1 · answered by Anqi 2 · 0⤊ 2⤋
248
Here's how to do it for 1-10
Divisible by?:
1: Of course. All integers are divisible by 1.
2: Yes. Last digit is 8, which is even.
3: No. Sum of the digits is 14, which isn't divisible by 3.
4: Yes. The last two digits are 48, which is divisible by 4.
5: No. The last digit is not a 0 or 5.
6: No. It is not divisible by both 2 and 3.
7: No. 248 - 3(70) = 248 - 210 = 38. 38 - 5(7) = 38 - 35 = 3, which is not divisible by 7. So it is not divisible by 7.
8: Yes. 248 - 3(80) = 248 - 240 = 8, which is divisible by 8. So it is divisible by 8.
9: No. The sum of the digits is 14, which is not divisible by 9.
10: No. The last digit is not a 0.
So the number is divisible by 2, 4, and 8.
2007-08-28 11:26:18 · answer #2 · answered by whitesox09 7 · 2⤊ 1⤋
1
248 = 2*2*2*31
1827 = 3*3*7*29
1570 = 2*5*157
348 = 2*2*3*29
no single common factor for all 4 numbers.
2007-08-28 11:25:00 · answer #3 · answered by Carborane 6 · 1⤊ 2⤋
Which of those numbers are divisible? All the even ones, for sure, they can all be divided by two, at the very least. Numbers ending in 7 can be divided by 7, 3, OR 9, in this case it looks like 1827 can at least be divided by 7, see if you can divide it by 3 and 9, too, maybe you can. Maybe I didn't understand your question, but it seems like all of them are divisible by something.
2007-08-28 11:25:04 · answer #4 · answered by jxt299 7 · 0⤊ 3⤋
248
2007-08-28 11:23:12 · answer #5 · answered by sdurio 2 · 0⤊ 3⤋
google calculator
2007-08-28 11:23:00 · answer #6 · answered by Ian H 2 · 1⤊ 2⤋